Physics

Electromagnetic Waves and Spectrum

659 Questions

Electromagnetic waves and spectrum questions test a candidate's understanding of radiation frequencies, wavelengths, and the properties of different rays like infrared, ultraviolet, and visible light. Concepts also cover practical applications in astronomy and the fundamental speed of light calculations. This physics topic appears regularly in general science sections of major competitive examinations.

UV rays propertiesElectromagnetic radiation frequencyWavelength identificationSpeed of light calculationsBlack body radiation

Electromagnetic Waves and Spectrum Questions

Multiple choice botany photosynthesis action spectrum and absorption spectrum spectrum of electromagnetic radiation chloroplast and pigments of photosynthesis site of photosynthesis

Which of the is the range for UV in spectroscopy?

  1. 400-700nm

  2. 500-800nm

  3. 200-300nm

  4. All of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

UV spectroscopy (400-700nm) is a type of absorption spectroscopy in which light of ultra-violet region is absorbed by the molecule resulting in the excitation of the electrons from the ground state to higher energy state. It obeys the Beer-Lambert law that states that when a beam of monochromatic light passes through a solution of an absorbing substance, the rate of decrease of intensity of radiation with the thickness of the absorbing solution is directly proportional to the incident radiation as well as the concentration of the solution.

So, the correct answer is '400-700nm'

Multiple choice botany photosynthesis action spectrum and absorption spectrum spectrum of electromagnetic radiation chloroplast and pigments of photosynthesis

Electromagnetic radiation with wavelengths between 0.4 and 0.7 micrometers is called:

  1. Ultraviolet light

  2. Visible light

  3. Infrared light

  4. Microwaves
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The visible spectrum of electromagnetic radiation ranges from approximately 0.4 micrometers (violet) to 0.7 micrometers (red).

Multiple choice limitations of bohr model and explanation of bohr's second postulate by matter waves bohr's model atoms atomic nuclei physics

Wavelength of first line in Lyman series is $\lambda $. The wavelength of first line in Balmer series is:

  1. $\dfrac { 5 }{ 27 } \lambda $
  2. $\dfrac { 32}{ 27 } \lambda $
  3. $\dfrac { 27 }{ 5 } \lambda $
  4. $\dfrac { 27 }{ 32 } \lambda $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

According to Bohr, the wavelength emitted when an electron jumps from ${ n } _{ 1 }^{ th }$ to ${ n } _{ 2 }^{ th }$ orbit is
$E=\dfrac { hc }{ \lambda  } ={ E } _{ 2 }-{ E } _{ 1 }$
$\Rightarrow \dfrac { 1 }{ \lambda  } =R\left( \dfrac { 1 }{ { n } _{ 1 }^{ 2 } } -\dfrac { 1 }{ { n } _{ 2 }^{ 2 } }  \right) $
For first line in Lyman series
$\dfrac { 1 }{ { \lambda  } _{ L } } =R\left( \dfrac { 1 }{ { 1 }^{ 2 } } -\dfrac { 1 }{ { 2 }^{ 2 } }  \right) =\dfrac { 3R }{ 4 } $                        ......(i)
For first line in Balmer series,
$\dfrac { 1 }{ { \lambda  } _{ B } } =R\left( \dfrac { 1 }{ { 2 }^{ 2 } } -\dfrac { 1 }{ { 3 }^{ 2 } }  \right) =\dfrac { 5R }{ 36 } $                      ......(ii)
From equations (i) and (ii)
$\therefore \dfrac { { \lambda  } _{ B } }{ { \lambda  } _{ L } } =\dfrac { 3R }{ 4 } \times \dfrac { 36 }{ 5R } =\dfrac { 27 }{ 5 } $
$\therefore { \lambda  } _{ B }=\dfrac { 27 }{ 5 } \lambda $               $\left( \because { \lambda  } _{ L }=\lambda  \right) $

Multiple choice limitations of bohr model and explanation of bohr's second postulate by matter waves bohr's model atoms atomic nuclei physics

If photon energy $E$ and an electron have same energy $E$ (kinetic energy) and De-Broglie wavelength of an electron is $\lambda _{e}$ and De-Broglie wavelength of photon is $\lambda _{p}$. The correct relation between $\lambda _{e}$ and $\lambda _{p}$ is 

  1. $\lambda _{p} \propto \lambda _{e}$
  2. $\lambda _{p} \propto \sqrt{\lambda _{e}}$
  3. $\lambda _{p} \propto \dfrac{1}{\sqrt{\lambda _{e}}}$
  4. $\lambda _{p} \propto \lambda _{e}^{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For photon

$E = \dfrac{{hc}}{{{\lambda _p}}}$---------------$(1)$
$ \Rightarrow {\lambda _p} = \dfrac{{hc}}{E}$

$E = \dfrac{1}{2}m{v^2}$

$ \Rightarrow v = \sqrt {\dfrac{{2E}}{m}} $

${\lambda _e} = \dfrac{h}{{mv}} = \dfrac{h}{{n\sqrt {\dfrac{{2E}}{m}} }} = \frac{h}{{\sqrt {2Em} }}$

$ \Rightarrow {\lambda _e}^2 = \dfrac{{{h^2}}}{{2Em}}$

$ \Rightarrow E = \dfrac{{{h^2}}}{{2m{\lambda _e}^2}}$
$\therefore \dfrac{{{h _e}}}{{{\lambda _p}}} = \dfrac{{{h^2}}}{{2m{\lambda _e}^2}}$

$ \Rightarrow {\lambda _p}\alpha \,{\lambda _e}^2$
Hence,
option $(D)$ is correct answer.

Multiple choice physics light and shadow formation of image by a pinhole camera pinhole camera shadow

Which one statement is correct -

  1. Speed of light in free space $= \frac{1}{\sqrt{\mu _0 \varepsilon _0}}$
  2. Speed of light in any medium $=\frac{1}{\sqrt{\mu \varepsilon}}$
  3. $\frac{E _0}{B _0}=2C$
  4. $\frac{B _0}{E _0}=4C$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The speed of light in free space is given by c = 1 / sqrt(mu_0 * epsilon_0), which is a fundamental result of Maxwell's equations.

Multiple choice laws of heat transfer heat and thermodynamics physics

Two black metallic spheres of radius 4m, at 2000 K and 1m at 4000 K will have ratio of energy radiation as

  1. 1 : 1

  2. 4 : 1

  3. 1 : 4

  4. 2 : 1
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Power P = sigma * A * T^4. P1 = sigma * 4 * pi * (4)^2 * (2000)^4. P2 = sigma * 4 * pi * (1)^2 * (4000)^4. Ratio P1/P2 = (16 * 2000^4) / (1 * 4000^4) = 16 / 16 = 1. The ratio is 1:1.

Multiple choice laws of heat transfer heat and thermodynamics physics

Choose the correct answer from the alternatives given.
Radiations of intensity $0.5\ W/m^2$ are striking a metal plate. The pressure on the plate is then

  1. $0.166 \, \times \, 10^{-8} \, N \, m^{-2}$
  2. $0.332 \, \times \, 10^{-8} \, N \, m^{-2}$
  3. $0.111 \, \times \, 10^{-8} \, N \, m^{-2}$
  4. $0.083 \, \times \, 10^{-8} \, N \, m^{-2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given: The intensity of the incident radiations is $0.5\ W/m^2$.


To find: The radiation pressure on the plate.

The radiation pressure experienced by the plate is given as;
$P = \dfrac{I}{c}\\Rightarrow \dfrac{0.5}{3 \times 10^8}\\Rightarrow  0.166 \times 10^{-8} N m^{-2}$

Option $(A)$ is correct.

Multiple choice laws of heat transfer heat and thermodynamics physics

Star A emits radiation of maximum intensity at a wavelength of $5000 \mathring{A}$ and it has temperature $ 1227^oC $. If star B has temperature $ 2727^oC $ , then the maximum intensity would be observed at 

  1. $ 4000 \mathring{A} $
  2. $ 2250 \mathring{A} $
  3. $ 3000 \mathring{A} $
  4. $ 2500 \mathring{A} $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\begin{array}{l} \dfrac { { \lambda { m^{ 1 } } } }{ { \lambda m } } =\dfrac { T }{ { { T^{ 1 } } } }  \ \dfrac { { \lambda { m^{ 1 } } } }{ { 5000A } } =\dfrac { { 1227+273k } }{ { 2727+273k } }  \ \lambda { m^{ 1 } }=2500A \end{array}$

Multiple choice electromagnetic spectrum electromagnetic waves physics

If $c$ is the speed, $\nu$ is frequency and $\displaystyle \lambda $ is wavelength of EM waves, then

  1. $\displaystyle c=\nu\lambda $
  2. $\displaystyle \frac { \lambda }{ \nu } =c$
  3. $\displaystyle \frac { \nu }{ \lambda } =c$
  4. $\displaystyle \frac { 1 }{ \lambda } =\frac { c }{ \nu} $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\displaystyle c=\nu \lambda $
$c=$speed
$\nu=$Frequency
$\displaystyle \lambda $ = wavelength

Multiple choice electromagnetic spectrum electromagnetic waves physics

The broad wavelength range of visible spectrum is:

  1. $4000-8000A^o$
  2. $2000-4000A^o$
  3. $10000-20000A^o$

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The wavelength range of $4000-8000 A^o$ is known as visible spectrum as waves within this wavelength range create a sensation of vision in our eyes.

Multiple choice electromagnetic spectrum electromagnetic waves physics

Identify which of the following light rays has the highest energy?

  1. Violet

  2. Green

  3. Yellow

  4. Orange

  5. Red
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Energy of light ray      $E = h\nu$                $\implies E \propto \nu$

Among all the visible rays, violet ray has the highest frequency, Thus violet ray has the highest energy. 

Multiple choice electromagnetic spectrum electromagnetic waves physics

The portion of the spectrum beyond the red end is called

  1. UV spectrum

  2. Infrard spectrum

  3. Microwave

  4. All
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The portion of spectrum just beyond the red end is called infrared spectrum, while the portion of the spectrum just before the voilet end is called the ultravoilet spectrum.

Multiple choice electromagnetic spectrum electromagnetic waves physics

Wavelength of gamma rays are :

  1. ${ 10 }^{ -10 }m$ to less than ${ 10 }^{ -14 }m$
  2. ${ 10 }^{ -14 }m$ to less than ${ 10 }^{ -10 }m$
  3. ${ 10 }^{ -11 }m$ to less than ${ 10 }^{ -14 }m$
  4. ${ 10 }^{ -14 }m$ to less than ${ 10 }^{ -6 }m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Electromagnetic radiations are present all around us in different forms such as microwaves, radio waves, gamma rays, and X-rays. These radiations can be defined as a form of energy produced by the movement of electrically charged particles that can be found in matter or vacuum or by oscillating magnetic or electric disturbance.


Properties of electromagnetic radiations :

1) They travel through empty space
2) The speed of light always remains constant i.e. 2.99792458 X 10 8 m/s.
3) Wavelength is the measure between the distance of either troughs or crests.  Its symbol is 'Lambda'.

Gamma rays have no mass.  They arise from the high-frequency end of the electromagnetic spectrum.  They have the highest penetration power.   They are at least ionizing.  The Gamma rays carry a large amount of energy and can travel through the thick and thin material. Gamma rays have frequencies greater than about 1018 cycles per second or Hertz.  They have wavelengths of less than 100 picometer.  Gamma rays can kill living cells.  It is used to kill cancerous cells.   They can also kill bacteria.

Multiple choice electromagnetic spectrum electromagnetic waves physics

Electromagnetic wave with frequencies greater than the critical frequency of ionosphere cannot be used for communication using sky wave propagation because

  1. The refractive index of ionosphere becomes very high for $f > f _ { c }$
  2. The refractive index of ionosphere becomes very low for $f > f _ { c }$
  3. The refractive index of ionosphere becomes very high for $f < f _ { c }$
  4. The refractive index of ionosphere becomes very low for $f < f _ { c }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The refractive index of the ionosphere is given by n = sqrt(1 - (f_c/f)^2). For f > f_c, the refractive index is real and less than 1, but as f increases, the ionosphere becomes less effective at refracting the wave back to Earth, eventually allowing it to pass through.